\(\left(x-y\right)^2\ge0;\forall xy\Rightarrow x^2+y^2\ge2xy\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow x+y\ge2\sqrt{xy}\)

\(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{2}{xy}\Rightarrow xy\ge4\Rightarrow x+y\ge2\sqrt{xy}\ge2\sqrt{4}=4\)

\(C_{min}=4\) khi \(x=y=2\)

Hoặc là:

\(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(\dfrac{4}{x+y}\right)^2=\dfrac{8}{\left(x+y\right)^2}\)

\(\Rightarrow\left(x+y\right)^2\ge16\Rightarrow x+y\ge4\)

Đề bài sai, C không có giá trị nhỏ nhất

Nếu \(\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{2}\) thì có thể tìm được min của C

Áp dụng BĐT AM-GM ta có:

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{\left(1+4\right)^2}{2}=\frac{5^2}{2}=\frac{25}{2}\)

Xảy ra khi \(x=y=\frac{1}{2}\)

\(P=\left(2x+\dfrac{1}{x}\right)^2+9+\left(2y+\dfrac{1}{y}\right)^2+9-18\)

\(P\ge2\sqrt{9\left(2x+\dfrac{1}{x}\right)^2}+2\sqrt{9\left(2y+\dfrac{1}{y}\right)^2}-18\)

\(P\ge12x+12y+\dfrac{6}{x}+\dfrac{6}{y}-18\)

\(P\ge6\left(4x+\dfrac{1}{x}\right)+6\left(4y+\dfrac{1}{y}\right)-12\left(x+y\right)-18\)

\(P\ge6.2\sqrt{\dfrac{4x}{x}}+6.2\sqrt{\dfrac{4y}{y}}-12.1-18=18\)

\(P_{min}=18\) khi \(x=y=\dfrac{1}{2}\)

Tham khảo: Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học trực tuyến OLM

\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)

Áp dụng BĐT Caushy-Schwarz ta có:

\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)

\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)

Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có 

\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)

\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)

    \(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)